How do I solve for $t$ in this equation?

Question

I know I'm supposed to use $\ln()$ to work it out, but I can't remember how it's done. Can anyone help?

The equation is

$$ 40e^{-t/5}=20 $$

Answer 1

You have $e^{-t/5}=1/2$ taking logaritms $-t/5=-\ln 2$.

Answer 2

$$40 e^{-t/5} = 20$$ $$ e^{-t/5} = \frac{1}{2}$$

Because of the definition of the natural logarithm function $ln(e^x)=x$, so you get:

$$-\frac{t}{5} = ln\left(\frac{1}{2}\right)$$

Considering that $ln(1/x) = - ln(x)$ you get to the desired answer.

$$ t = 5 ln(2)$$

Answer 3

Notice, $$40e^{-t/5}=20$$ $$e^{-t/5}=\frac{20}{40}=\frac{1}{2}$$ $$\ln e^{-t/5}=\ln \frac{1}{2}$$ $$-\frac{t}{5}=\ln (2^{-1})$$ $$-\frac{t}{5}=-\ln (2)$$ $$t=5\ln 2\approx 3.465735903$$

Answer 4

$$40e^{-t/5}=20$$ Divide both sides by $40$: $$e^{-t/5}= \frac{1}{2}$$ Take the natural logarithm: $$-t/5= -ln(2)$$ Multiply both sides by 5: $$-t= -5* ln(2)$$ And thus: $$t\approx 3.4657359027997265$$

Answer 5

$$ 40 \frac{1}{e^{t/5}} = 20 $$ You can divide away the factor $40$ and then calculate $\ln()$ of both sides $$ \ln\left(\frac{1}{e^{t/5}}\right) = \ln\left(\frac{1}{2}\right) $$ Then use a logarithm rule to move the "$-$" outside the $\ln()$ and obtain $t$: $$ \begin{align} -\ln\left(e^{t/5}\right) &= \ln\left(\frac{1}{2}\right)\\ -\frac{t}{5} &= \ln\left(\frac{1}{2}\right)\\ t &= -5\ln\left(\frac{1}{2}\right) \end{align} $$

Answer 6

You have $$40 \text{e}^{\frac{-t}{5}} = 20$$.

Divide both sides by $40$ to get $$\text{e}^{\frac{-t}{5}}=\frac{1}{2}$$

Now take the natural log of both sides to obtain $$\text{ln(e)}^{\frac{-t}{5}}=\text{ln}(\frac{1}{2})$$

Use the log operator to get $$-\frac{t}{5} \text{ln(e)}=\text{ln}(\frac{1}{2})$$

Note that $\text{ln(e)}=1$, so we obtain $$-\frac{t}{5}=\text{ln}(\frac{1}{2})$$.

Multiply both sides by $-5$ to obtain $$t=-5 \text{ln}\frac{1}{2} \approx 3.465735903.$$